Solving Systems of Linear Equations using the Personal Algebra Tutor illustrates the versatility of the program. The following example will be solved using five different methods.
The Personal Algebra Tutor:
Solution Using Elementary Operations
Y + X = 5 Equation # 1
Y - 2X = - 1 Equation # 2
| Next ...
| Subtract: Equation #2 - Equation #1 to eliminate Y
3X = 6 Eq. #1 and #2 Combined
| Next ...
| Divide by coefficient of variable to solve for X.
X = 2 Solution for X
| Next choose one of the original equations and
| substitute the value of X into that equation.
| ... Choosing Equation # 2
Y - 4 = - 1
Y = 3 Solution for Y
| **** Solution Set: X = 2, Y = 3,
Graphical Solution
Y + X = 5 Equation # 1
Y - 2X = - 1 Equation # 2
Plotting Y+X =5
Plot the X-Intercept(Red): Set y=0; Solve for x: (5,0)
Plot the Y-Intercept(Blue): Set x=0; Solve for y: (0,5)
Draw the line between the points.
Finding Slope-Intercept Equation of Line.
Solve for 'Y'.
Y= -1 X +5 Slope_Intercept Equation
Slope = -1 The coefficient of the X term.
Y-Intercept = 5 The constant term.
Done.
Plotting 2X - Y = 1
Plot the X-Intercept(Red): Set y=0; Solve for x: (0.5,0)
Plot the Y-Intercept(Blue): Set x=0; Solve for y: (0,-1)
Draw the line between the points.
Finding Slope-Intercept Equation of Line.
Solve for 'Y'.
Y= 2 X -1 Slope_Intercept Equation
Slope = 2 The coefficient of the X term.
Y-Intercept = -1 The constant term.
Done.
| The lines intersect at x=2, y=3
| The solution is x=2, y=3
***** Problem Summary *****
Problem
Y+X =5 Equation #1
2X - Y = 1 Equation #2
Solution
Solution: x=2, y=3
Solution Using Determinants
Y+X =5 Eq 1
2X-Y=1 Eq 2
Evaluate the Coefficient Determinant, D
| 1 1 |
D = | |
| 2 -1 |
D = 1*(-1) -(1)*(2) = -3
| 5 1 |
Dx= | |
| 1 -1 |
Dx= 5*(-1) -(1)*(1) = -6
Matrix Solution
Dx -6
X = ---- = ---- = 2
D -3
| 1 5 |
Dy= | |
| 2 1 |
Dy= 1*(1) -(5)*(2) = -9
Dy -9
Y = ---- = ---- = 3
D -3
**** X = 2, Y = 3 Solution is Consistent.
Matrix Solution Using Gauss Elimination and Back Substitution
Y+X =5 Eq 1
2X-Y=1 Eq 2
| 1 1 |
Mc= | 2 -1 | Coefficient Matrix
| 1 1 | 5 |
Ma= | 2 -1 | 1 | Augmented Matrix
Solve using the Augmented Matrix
| 1 1 | 5 |
M = | 0 3 | 9 | 2 R1 - R2 -> R2
Force all Row leading non-zero coefficients to 1.
| 1 1 | 5 | X +1 Y = 5
M = | 0 1 | 3 | Y = 3
Back Substitution
Y = 3
X = 5 -(1)Y = 5 -(1)(3)
X = 2
**** X = 2, Y = 3 Solution is Consistent.
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Matrices Using Gauss-Jordan Elimination
Y+X =5 Eq 1
2X-Y=1 Eq 2
| 1 1 |
Mc= | 2 -1 | Coefficient Matrix
| 1 1 | 5 |
Ma= | 2 -1 | 1 | Augmented Matrix
Solve using the Augmented Matrix
| 1 1 | 5 |
M = | 0 3 | 9 | 2 R1 - R2 -> R2
| -3 0 | -6 | R2 - 3 R1 -> R1
M = | 0 3 | 9 |
Force all Row leading non-zero coefficients to 1.
| 1 0 | 2 | X = 2
Mj= | 0 1 | 3 | Y = 3
**** X = 2, Y = 3 Solution is Consistent.
System of 3 Linear Equations Example Matrices Using Gauss-Jordan Elimination
2X-Y+Z=44 Eq 1
-X+3Y-2Z=-53 Eq 2
5X-6Y-Z=19 Eq 3
| 2 -1 1 |
Mc= | -1 3 -2 | Coefficient Matrix
| 5 -6 -1 |
| 2 -1 1 | 44 |
Ma= | -1 3 -2 | -53 | Augmented Matrix
| 5 -6 -1 | 19 |
Solve using the Augmented Matrix
| 2 -1 1 | 44 |
M = | -1 3 -2 | -53 |
| 0 7 7 | 182 | 5 R1 - 2 R3 -> R3
| 2 -1 1 | 44 |
M = | 0 5 -3 | -62 | R1 + 2 R2 -> R2
| 0 7 7 | 182 |
| 2 -1 1 | 44 |
M = | 0 5 -3 | -62 |
| 0 0 -56 | -1344 | 7 R2 - 5 R3 -> R3
| -112 56 0 | -1120 | - R3 + 56 R1 -> R1
M = | 0 5 -3 | -62 |
| 0 0 -56 | -1344 |
| -112 56 0 | -1120 |
M = | 0 280 0 | 560 | -3 R3 + 56 R2 -> R2
| 0 0 -56 | -1344 |
| -112 56 0 | -1120 | R1/280 -> R1
M = | 0 1 0 | 2 |
| 0 0 -56 | -1344 |
| 112 0 0 | 1232 | 56 R2 - R1 -> R1
M = | 0 1 0 | 2 |
| 0 0 -56 | -1344 |
Force all Row leading non-zero coefficients to 1.
| 1 0 0 | 11 | X = 11
Mj= | 0 1 0 | 2 | Y = 2
| 0 0 1 | 24 | Z = 24
**** X = 11, Y = 2, Z = 24 Solution is Consistent.
